Chemistry • Year 12 • Module 5 • Lesson 12
Reaction Quotient Q
Apply the Q vs Keq framework to real concentration data, graphs, disturbance scenarios and Australian industrial contexts.
1. Interpret concentration data — H2 + I2 ⇌ 2HI at 430 °C
For the reaction H2(g) + I2(g) ⇌ 2HI(g), Keq = 54.3 at 430 °C. The table below shows five different reaction mixtures (not at equilibrium) in a sealed vessel. 8 marks
| Mixture | [H2] (mol/L) | [I2] (mol/L) | [HI] (mol/L) | Q (calculate) | Q vs Keq | Direction of shift |
|---|---|---|---|---|---|---|
| A | 0.10 | 0.10 | 0.00 | |||
| B | 0.00 | 0.00 | 0.30 | |||
| C | 0.20 | 0.20 | 0.93 | |||
| D | 0.05 | 0.05 | 0.09 | |||
| E | 0.020 | 0.020 | 0.148 |
1.1 Complete the table by calculating Q for each mixture, comparing it to Keq = 54.3, and stating the direction of shift. Show working for Mixture C below. 5 marks
1.2 Identify which mixture (A, B, C, D or E) is already at equilibrium. Justify your answer with reference to Q and Keq. 2 marks
1.3 For Mixture B, explain in one sentence what Q = ∞ indicates about the starting conditions. 1 mark
2. Graph interpretation — Q approaching Keq over time
The graph below models how the reaction quotient Q changes over time as the system H2(g) + I2(g) ⇌ 2HI(g) approaches equilibrium from two different starting conditions. Keq = 54.3 at 430 °C. 8 marks
Stylised model for H₂ + I₂ ⇌ 2HI at 430 °C (Kₑₓ = 54.3). Time axis is qualitative.
2.1 Describe the trend in Q for the blue (solid) curve between time 0 and equilibrium. State what happens to H2, I2 and HI concentrations to cause this trend. 3 marks
2.2 Both curves converge on the same Q value at equilibrium, even though they started from opposite extremes. Explain why this must be the case. 2 marks
2.3 Estimate (from the graph) the approximate time at which each mixture reaches equilibrium. At this point, what is the value of Q for both mixtures? 2 marks
2.4 A student claims: “The red curve (starting from products) reaches equilibrium faster because it has further to travel.” Identify the error in this reasoning. 1 mark
3. Cause-and-effect chain — adding NH3 to the Haber process
The Haber process at Incitec Pivot’s Gibson Island (Brisbane) plant runs the reaction N2(g) + 3H2(g) ⇌ 2NH3(g). Suppose the system is at equilibrium and an operator adds extra NH3 to the reactor. Use Q reasoning to complete the cause-and-effect chain below. 6 marks
Effect on Q expression (1 mark)
New Q vs Keq (1 mark)
Direction of shift (1 mark)
What happens to [NH3], [N2], [H2] (1 mark)
What happens to Q as the system moves to the new equilibrium? (1 mark)
Overall outcome for the operator (LCP consistent?) (1 mark)
4. Case study — phosphate saturation in the Murray–Darling Basin
Read the passage, then answer the question below. 5 marks
Water quality scientists monitoring phosphate levels in the Murray–Darling river system use Q (known in water chemistry as the ion activity product, IAP) to assess whether calcium phosphate is precipitating or dissolving in river sediments. The relevant equilibrium is:
Ca3(PO4)2(s) ⇌ 3Ca2+(aq) + 2PO43−(aq), Ksp = 2.07 × 10−33
After a heavy-rainfall event washes fertiliser runoff into the Darling River, scientists measure [Ca2+] = 4.0 × 10−3 mol/L and [PO43−] = 6.5 × 10−7 mol/L. They use Q (= IAP = [Ca2+]3[PO43−]2) to decide whether to intervene to reduce phosphate loading. If Q > Ksp, calcium phosphate will precipitate, removing phosphate from solution; if Q < Ksp, no precipitation occurs and dissolved phosphate remains available to fuel algal blooms.
4.1 Calculate Q (IAP) for the river water described above. Show full working. 2 marks
4.2 Compare Q to Ksp and predict whether calcium phosphate will precipitate under these conditions. State what this means for the risk of an algal bloom. 2 marks
4.3 In one sentence, explain why monitoring Q rather than just measuring [PO43−] alone gives scientists more useful information about the river’s phosphate dynamics. 1 mark
5. Predict and justify — diluting an equilibrium mixture
4 marks
A sealed flask contains an equilibrium mixture of H2(g), I2(g), and HI(g) at 430 °C with Keq = 54.3. The volume of the flask is suddenly doubled (all concentrations are halved simultaneously). The reaction is H2(g) + I2(g) ⇌ 2HI(g).
Using the Q expression, predict whether the equilibrium will shift left, shift right, or remain unchanged after the volume doubles. Justify your prediction by calculating the new Q in terms of the original equilibrium concentrations and comparing it to Keq. Show your reasoning algebraically.
Q1 — Concentration data table
A: Q = 0²/(0.10×0.10) = 0. Q < Keq. Shift right.
B: Q = 0.30²/(0×0) = ∞ (denominator zero). Q > Keq. Shift left.
C: Q = 0.93²/(0.20×0.20) = 0.8649/0.0400 = 21.6. Q < Keq = 54.3. Shift right. Working sample: numerator = (0.93)² = 0.8649; denominator = (0.20)(0.20) = 0.0400; Q = 0.8649/0.0400 = 21.6.
D: Q = 0.09²/(0.05×0.05) = 0.0081/0.0025 = 3.24. Q < Keq. Shift right.
E: Q = 0.148²/(0.020×0.020) = 0.021904/0.000400 = 54.8 ≈ 54.3. Q = Keq (within rounding). At equilibrium — no net shift.
1.2: Mixture E is at equilibrium; Q ≈ 54.3 = Keq within experimental rounding, so no net shift occurs.
1.3: Q = ∞ means no reactants (H2 and I2) are present—the mixture consists of pure products only (HI), so the denominator of Q is zero and Q is mathematically infinite; the system must shift left to form reactants.
Q2 — Graph interpretation
2.1: The blue (solid) curve rises from Q = 0 at time 0, initially steeply then with decreasing gradient, levelling off asymptotically at Keq = 54.3. This occurs because the reaction shifts right: [HI] increases (numerator of Q increases) while [H2] and [I2] decrease (denominator decreases), so Q increases progressively. [3 marks: 1 describe shape; 1 HI increases numerator; 1 H2/I2 decrease denominator]
2.2: Both curves converge on Keq = 54.3 because Keq is a constant at a fixed temperature (430 °C) and represents the single equilibrium composition the system tends toward regardless of starting conditions. [1 Keq is constant at fixed T; 1 equilibrium is the same destination regardless of starting point]
2.3: From the graph, both mixtures reach approximate equilibrium at around time 5–6 (arbitrary units). At that point Q = 54.3 for both mixtures.
2.4: The student is wrong because the rate of approach to equilibrium depends on reaction kinetics (concentrations, temperature, activation energy), not on how far Q is from Keq. “Further to travel” (in Q-units) does not mean faster in real time; both mixtures can reach equilibrium at similar times depending on rate constants.
Q3 — Cause-and-effect chain: adding NH3
Effect on Q expression: NH3 is in the numerator of Q (= [NH3]²/([N2][H2]³)). Adding NH3 increases [NH3] → numerator increases → Q increases.
New Q vs Keq: Q > Keq.
Direction of shift: Shift left (reverse direction).
Concentrations: [NH3] decreases; [N2] and [H2] increase as NH3 decomposes.
Q as system shifts: Q decreases back toward Keq as [NH3] falls and [N2]/[H2] rise, until Q = Keq at the new equilibrium.
Overall outcome: The shift is consistent with Le Chatelier’s Principle—the system opposes the increase in NH3 by converting it back to reactants. The final [NH3] is higher than the original but lower than immediately after addition. [N2] and [H2] are slightly higher than their original values.
Q4 — Murray–Darling phosphate case study
4.1 Calculation:
Q = [Ca2+]³ × [PO43−]²
= (4.0 × 10−3)³ × (6.5 × 10−7)²
= (6.4 × 10−8) × (4.225 × 10−13)
= 2.7 × 10−20 [2 marks: 1 correct substitution; 1 correct answer with units noted as mol5/L5]
4.2: Q = 2.7 × 10−20 > Ksp = 2.07 × 10−33. Therefore Q > Ksp → the solution is supersaturated with respect to calcium phosphate → precipitation will occur. This is good news for algal bloom risk: as Ca3(PO4)2 precipitates, dissolved [PO43−] is reduced, limiting the nutrient available to fuel algal growth. [1 Q > Ksp stated; 1 precipitation → reduced dissolved phosphate → lower bloom risk]
4.3: Q incorporates both [Ca2+] and [PO43−] together in one expression, so it captures the combined effect of all relevant ions on whether precipitation is thermodynamically favoured, whereas monitoring [PO43−] alone cannot predict whether precipitation will actually occur without knowing [Ca2+] as well.
Q5 — Dilution of H2 + I2 ⇌ 2HI
Let the original equilibrium concentrations be [H2] = a, [I2] = b, [HI] = c (where Q = c²/(ab) = 54.3).
After doubling volume, all concentrations halve: [H2] = a/2, [I2] = b/2, [HI] = c/2.
New Q = (c/2)² / ((a/2)(b/2)) = (c²/4) / (ab/4) = c²/ab = 54.3.
Because the stoichiometry is 1:1:2 and the number of moles of gas is equal on both sides (1 + 1 = 2), the (1/2) factors cancel identically in numerator and denominator. Therefore the new Q = original Q = Keq. The equilibrium does not shift. This is unique to reactions where Δn(gas) = 0; for reactions where Δn ≠ 0, dilution would shift the equilibrium toward the side with more moles of gas.